Precalculus (6th Edition)

Published by Pearson
ISBN 10: 013421742X
ISBN 13: 978-0-13421-742-0

Chapter 1 - Equations and Inequalities - 1.8 Absolute Value Equations and Inequalities - 1.8 Exercises - Page 167: 33


The solution is $(-\infty,0)\cup(6,\infty)$

Work Step by Step

$4|x-3|\gt12$ Take $4$ to divide the right side: $|x-3|\gt\dfrac{12}{4}$ $|x-3|\gt3$ Solving this absolute value inequality is equivalent to solving two separate inequalities, which are: $x-3\gt3$ and $x-3\lt-3$ $\textbf{Solve the first inequality:}$ $x-3\gt3$ Take $3$ to the right side: $x\gt3+3$ $x\gt6$ $\textbf{Solve the second inequality:}$ $x-3\lt-3$ Take $3$ to the right side: $x\lt-3+3$ $x\lt0$ Expressing the solution in interval notation: $(-\infty,0)\cup(6,\infty)$
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